$\overline{AB} = 10$ $\overline{AC} = {?}$ $A$ $C$ $B$ $10$ $?$ $ \sin( \angle BAC ) = \frac{3}{5}, \cos( \angle BAC ) = \frac{4}{5}, \tan( \angle BAC ) = \dfrac{3}{4}$
Answer: $\overline{AB}$ is the hypotenuse $\overline{AC}$ is adjacent to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the adjacent side so we can use the cos function (CAH) $ \cos( \angle BAC ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\overline{AC}}{\overline{AB}}= \frac{\overline{AC}}{10} $ $ \overline{AC}=10 \cdot \cos( \angle BAC ) = 10 \cdot \frac{4}{5} = 8$